# Find the explicit formula for the nth term of the sequence?

1. Apr 3, 2013

### Topgun_68

1. The problem statement, all variables and given/known data

Consider the sequence n1, n2, n3, ... that satisfies the recurrence relation nk = nk-1 / k + 1 for all integers k ≥ 2 with the initial condition that n1 = 1. Find the explicit formula nk for the nth term of the sequence?

2. The attempt at a solution

I calculated out the terms for n = 1, n = 2, n = 3 & n = 4 but there is no obvious relation that I can see because of the decimal numbers. For example:

Starting variable n1 = 1 so..

n2 = $\frac{1}{3}$ = .333

n3 = $\frac{.333}{4}$ = .08325

n4 = $\frac{.08325}{5}$ = .01665

Any hints on how to calculate the explicit formula for nk?
Thanks for any info!

2. Apr 3, 2013

### SteamKing

Staff Emeritus
What do you notice about the relationship between the denominators and the index k?
What relationship do you deduce about the numerators and n-sub-k?

3. Apr 3, 2013

### Topgun_68

Hmm, the denominators are increasing sequentially by 1 more than k.

As the index goes up the numerator is getting smaller.

I'll keep playing with the number until something jumps out at me. I don't know where my instructor comes up with these tough problems, but I can do the easy ones in the book

4. Apr 3, 2013

### HallsofIvy

Staff Emeritus
It would have been better NOT to approximate with a decimal

n3= (1/3)/(3+1)= 1/(3(4))/

n4= (1/(3(4)))/(4+ 1)= 11/(3(4)(5))

1/3, 1/(3(4)), 1/(3(4)(5)), should remind you of a factorial.

5. Apr 3, 2013

### Topgun_68

Ah, I never would have noticed with the decimal numbers. I won't make that mistake again. It looks like the factorial starts at 3 so is he equation correct below? I get correct answers with a numerator of 2 but can't figure out how to get the 2? I just guessed it. Thanks for the help!

nk = $\frac{2}{(n+1)n!}$

6. Apr 3, 2013

### Ray Vickson

What you wrote means
$$n_k = \frac{n_{k-1}}{k} + 1.$$
Did you mean that, or did you mean
$$n_k = \frac{n_{k-1}}{k+1} ?$$
If you mean the latter, USE PARENTHESES, like this: nk = nk-1/(k+1).

Anyway, do not use limited decimal representations; 1/3 is NOT 0.333, 1/12 is NOT 0.8325 (it is actually .08333333333333333 ....) The errors you get are building up more and more as you increase the number of steps, and using decimals like that serves no useful purpose here.

7. Apr 3, 2013

### Mentallic

Since you have

$$n_k = \frac{1}{3.4.5...k}$$

and

$$\frac{1}{k!}=\frac{1}{2.3.4.5...k}=\frac{1}{2}\cdot\frac{1}{3.4.5...k}=\frac{1}{2}\cdot n_k$$

Does this manipulation make sense so far?

So finally, what is nk?

EDIT: Also, you guessed the formula to be

$$\frac{2}{(n+1)n!}$$

which I should note is equivalent to

$$=\frac{2}{(n+1)!}$$

And also, you're looking for the kth term, not the nth term. Your formula should be in terms of k.

$$n_k = f(k)$$

So if you're looking for n2 then you plug k=2 into the formula.

Last edited: Apr 3, 2013
8. Apr 4, 2013

### Topgun_68

Sorry, I am trying to learn how to put the equations in using the correct symbols, so I forgot the parenthesis, which I know totally changes the equation. I meant the later..

$$n_k = \frac{(n_{k-1})}{(k+1)} ?$$

Last edited: Apr 4, 2013
9. Apr 4, 2013

### Topgun_68

Thanks for all the clarifications. I have been mixing up the terms and it's been hurting my grades.

so would it be

$$n_k = \frac{2}{(k+1)!}$$

Thanks. That's what I like about this forums. You can always count on everyone to correct all these errors. I have been learning a lot from them

10. Apr 4, 2013

### Mentallic

No. Look back to my derivation. I ended up with

$$\frac{1}{k!}=\frac{1}{2}\cdot n_k$$

so then

$$n_k = \frac{2}{k!}$$

Did it make sense to you?

11. Apr 4, 2013

### Topgun_68

Yes I can see how you did all the math to acquire the answer, but how you got the $\frac{1}{2}$ confuses me a little. Is it because the question stated k ≥ 2 so you pulled it out of the factorial? Than you multiply both sides by 2 to isolate the nk.

Thanks for your patience with me!

12. Apr 4, 2013

### Mentallic

Sorry for misleading you, but I just wrote down the pattern for myself and noticed I did make a big blunder, you were right, it's

$$n_k = \frac{2}{(k+1)!}$$

My derivation in the earlier post still follows the same procedure, except that we begin with

$$n_k = \frac{1}{3.4.5...k(k+1)}$$

So then

$$\frac{1}{(k+1)!} =\frac{1}{2.3.4.5...k(k+1)} =\frac{1}{2} \frac{1}{3.4.5...k(k+1)} = \frac{1}{2}n_k$$

Hence we have that

$$n_k = \frac{2}{(k+1)!}$$

The factor of 1/2 came from the fact that when we noticed the pattern for nk to be

$$n_k = \frac{1}{3.4.5...k(k+1)}$$

The denominator is nearly

$$(k+1)! = 2.3.4.5...k(k+1)$$

But we're missing the 2, so we put in there! But if we multiplied the denominator by 2, then in order to keep it equal, we need to multiply the numerator by 2 as well:

$$\frac{1}{3.4.5...(k+1)}=\frac{2}{2}\cdot \frac{1}{3.4.5...(k+1)} = \frac{2}{2.3.4...(k+1)}=\frac{2}{(k+1)!}$$

If the question was instead changed to "for integers $k\geq 1$" then our initial condition would have to be on n0, so let n0=1, then

$$n_1=\frac{n_0}{1+1}=\frac{1}{2}$$

$$n_2=\frac{n_1}{2+1}=\frac{1/2}{3}=\frac{1}{2.3}$$

Following this pattern, we can see that kth term will be

$$n_k = \frac{1}{2.3.4...k(k+1)}$$

Which is simply

$$n_k = \frac{1}{(k+1)!}$$

13. Apr 4, 2013

### Topgun_68

Ah, that's where the two came from. It cancels out the almost factorial

Thanks again for everyone's assistance. I like how everyone on here helps you figure out the answer for yourself, which is turn makes it's easier to understand come exam time.